Homes in rural areas where natural gas service is not available often rely on propane to fuel kitchen ranges. The propane is stored as a liquid, and the gas to be burned is produced as the liquid evaporates. Suppose an architect has hired you to consult on the choice of a propane tank for such a new home. The propane gas consumed in 1.0 hour by a typical range burner at high power would occupy roughly 165 L at 25°C and 1.0 atm, and the range chosen by the client will have 4 burners. If the tank under consideration holds 400 gallons of liquid propane, what is the minimum number of hours it would take for the range to consume an entire tank of propane? The density of liquid propane is 0.5077 kg/L.

The key to solve this problem is that we are given the propane gas consumed in one hour by giving us the information of the volume consumed at 1 atm, 298 K (25 +273). Using the gas law we can calculate the rate of consumption of propane per hour, and from here we can calculate its mass and converting it to gallons and finally diving the 400 gallos by this number.

PV = nRT ∴ n = PV/RT

n = 1 atm x 165 L/ (0.08206 Latm/kmol x 298 K ) = 6.75 mol propane

Mass propane :

6.75 mol x 44 g/mol = 296.88 g

convert this to Kg:

296.88 g/ 1000 g/Kg = 0.30 Kg

calculate the volume in liters this represents by dividing by the density:

0.30 Kg / 0.5077 Kg/L = 0.59 L

changing this to gallons

0.59 L x 1 gallon/3.785 L = 0.15 gallon

and finally calculate how many hours the 400 gallons propane tank will deliver

400 gallon/ 0.15 gallon/hr = 2666.7 hr